4.3 The Process of Nondimensionalisation
4.3.1 Projectile Motion
Consider a projectile of mass Mkilograms that is launched vertically with initial speedV0m/s, from a positionY0meters above the surface of the Earth. Newton’s universal law of gravitation coupled with the second law of motion then gives that the height of the projectileY(T)varies with timeTaccording to the ordinary differential equation
Md2Y
dT2 = − GMEM
(RE+Y)2, Y(0)=Y0, Y(0)=V0, (4.2) where the Earth’s properties (mass and radius) areME = 6×1024kg and RE = 6.4×106m, and the universal gravitational constantG=6.7×10−11m3/(s2kg).
We begin by noting that the parametersGandMEonly occur as a product and so can only appear in the solution in the same form. Consequently, we can regard this combination as a single parameter and replace it usingg=GME/R2E≈9.81 m/s2, this being the familiar value for acceleration due to gravity at the Earth’s surface.
The problem then takes the form Md2Y
dT2 = − gR2EM
(RE+Y)2, Y(0)=Y0, Y(0)=V0. (4.3) We now introduce the (as yet arbitrary) length and time scales, L and T,
Y(T)=Ly(t), T=Tt, (4.4)
4.3 The Process of Nondimensionalisation 89 so thatyandtare dimensionless variables. The scaling constants (L and T) for the different independent dimensional quantities in a problem (here, length and time) are called thecharacteristic scales.
Substituting into (4.3) yields the rescaled problem L
T2 d2y
dt2 = −g 1
1+
L RE
y
2, Ly(0)=Y0, L
Ty(0)=V0.
Dividing through each expression by the dimensional coefficient of the first term on the left-hand-side yields thenondimensionalised problem
d2y dt2 = −
gT2 L
Π1
⎛ 1
⎜⎜
⎜⎝1+ L
RE Π2
y
⎞
⎟⎟
⎟⎠
2, y(0)= Y0
L Π3
, y(0)= V0T
L Π4
,
(4.5) where theΠiare nondimensional constants. It is important to realise that such dimen- sionless parameters always represent the ratio of two competing effects. For example, Π1represents the relative importance of gravitational acceleration versus the accel- eration of the projectile,Π1 =g/(L/T2). Likewise,Π2compares the length scale of interest to the Earth’s radius,Π2=L/RE.
We have the freedom to choose the length and time scales L and T to scale the problem into a desired form; here we describe one such form. Selecting the length scale L =Y0setsΠ3 =1, while choosing a timescale based on acceleration due to gravity, T = (L/g)1/2, fixesΠ1 = 1. Through (4.4), the selection of L and T gives the respective units of measure that all lengths and times in the problem will be calibrated against. Characteristics scales given directly by a single given quantity from the original problem (here L=Y0) are sometimes calledimposed scales. Scales obtained from combinations of given quantities (here T in terms ofY0andg) are calledderived scales.
We note that there are, in fact, many possible scaling choices as we could, for example, multiply either (or both) of the length or time scales by some function f(Y0/RE)and the resulting equation would still be nondimensional. However, it turns out that the normalised forms are often the most convenient choices.
We have employed what we shall refer to as
The 1st general nondimensional scaling principle:
Select the characteristic scales so that as many as possible of theΠ’s are normalised.
(4.6)
For the current example, the nondimensionalised model takes the simplified form d2y
dt2 = − 1
(1+Π2y)2, y(0)=1, y(0)=Π4. (4.7a) with
Π2= L RE = Y0
RE, Π4=V0T
L = V0
(gY0)1/2. (4.7b) The solution of (4.7a) will be a function of the independent variabletand the remain- ing nondimensional constantsΠ2andΠ4, namely
y=y(t, Π2, Π4), or equivalently,
Y=Y0y(t, Π2, Π4), T=(Y0/g)1/2t,
in contrast to the solution of the original problem (4.3) which takes the general form Y=Y(T,g,RE,Y0,V0).
We have effectively reduced the complexity of the original problem (involving a solution function of five variables) to a solution depending on only three variables.
As will be discussed further, nondimensionalization allows us to examine different limiting forms of problems in a mathematically precise framework. Consider the limit Π2 = Y0/RE → 0 in the model (4.7). At first sight, this limit may appear to describe increasing the radius of the Earth (RE→ ∞), or decreasing the initial height (Y0→0), or both. For the limitRE→ ∞(withY0fixed), we can formally setΠ2=0 to reduce (4.7a) to a simpler ODE that can be easily solved,
d2y
dt2 = −1 =⇒ y(t)=1+Π4t−12t2.
This solution can be interpreted as afirst approximationto the motion of a projectile that is launched relatively close to the surface of a large-radius planet with the same gravity as the Earth. Undoing the nondimensional scaling (4.4) converts this to the dimensional form of the classic quadratic polynomial for projectile motion,
Y=Y0+V0T−12gT2.
However if we consider the limitY0→0 (withREfixed) thenΠ4→ ∞; this is not acceptable as a value of for the initial condition on y(0)and forces us to question whether we have selected an appropriate choice of characteristic scales.
In fact, if scalings yield any undefined terms in a non-dimensionalized model, then that model in that form will not be solvable. This points to the need to change the scalings to put the model into a well-defined form, which we express as
4.3 The Process of Nondimensionalisation 91
The 2nd general nondimensional scaling principle:
Select characteristic scales so that NO terms in the model diverge in the physical limit of interest.
(4.8)
The choice of scales is determined once the sub-set of Π’s to be normalised is made; hence other choices correspond to picking differentΠ’s. For problems withk characteristic scales, the scales will be determined from normalisingkΠ’s; this will be discussed further in connection with the Buckingham Pi theorem, see Sect.4.5.
To examine theY0→0 limit, consider settingΠ1=1 andΠ4=1 in (4.5). The length and time scales are then L=V20/gand T =V0/g. The revised form of the nondimensionalised problem is
d2y
dt2 = − 1
(1+ ˜Π2y)2, y(0)= ˜Π3, y(0)=1. (4.9a) Π˜2= L
RE = V20
gRE, Π˜3= Y0
L = Y0g
V20 . (4.9b)
This choice of scalings satisfies (4.8) since settingY0=0 setsΠ˜3=0 but does not cause any other parameters in the problem to diverge,
d2y
dt2 = − 1
(1+ ˜Π2y)2, y(0)=0, y(0)=1.
If we further specify thatRE→ ∞thenΠ˜2=0 and the ODE reduces toy= −1 yielding the solution
y(t)= −12t2+t =⇒ Y=V0T−12gT2.
ForV0>0 thetime of flightof the projectile is clearly seen to be 2V0/gwhile the maximum height reached isV20/(2g), both of which are directly proportional to the scales that we chose.